It's one of the common indeterminate forms. Here are some more:
∞ - ∞
∞*0
∞/∞
0/0
0^0
∞^0
1^∞
#is -inf + inf = 0 or is it undetermiend
mint crow
alpine parcel
infinity is short hand for "unbounded from above"
could be anything
for instance
$$ \lim _{m\to\infty} m+1-m = 1\qquad \lim _{m\to\infty} m+2-m = 2 $$
frozen heraldBOT
aL
earnest bough
@alpine parcel wait so if you have two variables, both of which approach infinity then they cancel each other out?
Also you are contradicting what @mint crow is saying so im a bit confused
mint crow
Also you are contradicting what <@290136354233384960> is saying so im a bit conf...
That doesn't contradict what I'm saying.
What's important is that even if lim(g(x)) and lim(f(x)) are infinite, lim(f(x) + g(x)) might still be finite.
earnest bough
@mint crow alright i think i understand, f(x) - f(x) is always going to equate to 0 even if the limit of f(x) approaches infinity as x approaches a but when you have seperate variables it is impossible to tell what they evaluate to under an operation
mint crow
<@290136354233384960> alright i think i understand, f(x) - f(x) is always going...
Yes, that's correct!
earnest bough
what if you had m/m-1 as m approaches infinity
wiat
l hopitals rule right>
you could evaluate
mint crow
what if you had m/m-1 as m approaches infinity
That's ∞/∞. You can do the following to resolve this case:
m/(m - 1) = 1/(1 - 1/m)
mint crow
l hopitals rule right>
Try to use it as rarely as possible.
It's not a good thing to overuse it.
mint crow
That's ∞/∞. You can do the following to resolve this case:
m/(m - 1) = 1/(1 - 1/...
Or:
m/(m - 1) = 1 + 1/(m - 1)
mint crow
You're welcome!
mint crow
how do I close
Type +close.
mint crow
No need for the dot at the end.
earnest bough
+close